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(A) Let the given determinant be $\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ {a^2 - bc} & {b^2 - ac} & {c^2 - ab} \end{array}} ig

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(A) Let the given determinant be $\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ {a^2 - bc} & {b^2 - ac} & {c^2 - ab} \end{array}} ight|$.We can split the determinant into two parts:$\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{array}} ight| - \left| {\begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ bc & ac & ab \end{array}} ight|$.For the second determinant,multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$ and divide by $abc$:$\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{array}} ight| - \frac{1}{abc} \left| {\begin{array}{ccc} a & b & c \ a^2 & b^2 & c^2 \ abc & abc & abc \end{array}} ight|$.Taking $abc$ common from $R_3$ in the second determinant:$\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{array}} ight| - \frac{abc}{abc} \left| {\begin{array}{ccc} a & b & c \ a^2 & b^2 & c^2 \ 1 & 1 & 1 \end{array}} ight|$.By swapping rows to match the first determinant,we see the two determinants are identical.Thus,$\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{array}} ight| - \left| {\begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{array}} ight| = 0$.Therefore,$2 	imes \Delta = 2 	imes 0 = 0$.

$2\,\,\left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ {a^2 - bc} & {b^2 - ac} & {c^2 - ab} \end{array}} \right| = $

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